![[An equilateral triangle inscribed in a circle.]](circle3.gif)
![[A square inscribed in a circle.]](circle4.gif)
![[A regular pentagon inscribed in a circle.]](circle5.gif)
You should know by now that the area of a circle is given by the formula πr2, but how do we go about showing that this is the correct formula? One way is to consider regular polygons inscribed inside the circle like this:
![[A regular pentagon inscribed in a circle with a sector shaded.]](circle5s.gif)
Notice how, as the number of sides increases, the difference between the
area of the circle and the area of the polygon gets smaller and smaller.
We'd guess, then, that by finding the area of these regular polygons, we can
approximate the area of the circle. Which leads us to the
question of how we find the area of the polygons.
The trick is to break the polygon into triangular segments, like the one illustrated by the shaded area of the pentagon:
A regular polygon with n sides will have n such triangles. The angle at the centre of such a triangle will be 2π/n radians. A typical triangle looks something like this (for simplicity, we will assume that the radius of the circle is 1):
![[An isoceles triangle OAB with a perpendicular dropped from O to AB intersecting at C. Angle AOC is theta. OA has length 1, OC has length cos(θ) and AC has length sin(θ).]](triangle.gif)
| Sides | Area |
|---|---|
| 3 | 1.29903810568 |
| 4 | 2.0 |
| 5 | 2.37764129074 |
| 6 | 2.59807621135 |
| 8 | 2.82842712475 |
| 10 | 2.93892626146 |
| 100 | 3.13952597647 |
| 1000 | 3.14157198278 |
| 10000 | 3.14159244688 |
| 100000 | 3.14159265152 |
| 1000000 | 3.14159265357 |
| 1000000000 | 3.14159265359 |
From basic trigonometry, we can work out the lengths of the sides to be as shown, where θ = π/n radians. As a result, the area of the triangle must be
cos(π/n) sin(π/n)
And since n such triangles make up the area of a regular polygon, the total area of an n-sided polygon is
An = n cos(π/n) sin(π/n)
So now we can use this to approximate the area of the circle. For small numbers of sides, the approximations are not that good, but for large numbers of sides (such as a billion), the approximation to the real area (which is π units) is very good.
Recall that a tangent to a circle is a line which interects the circle in exactly one point. Given a circle centred at O and a point P, we'd like to find the equation of the tangent through P. We could solve this problem using some simple trigonometry or plane geometry, but instead we'll try another method which doesn't use any trigonometry.
Assume that we have a circle centred at O = (0,0), with radius 5 units, and we want to find the tangent at the point P = (3,4). Recall that the point-slope formula for the equation of a line tells us that the tangent line must be of the form:
y - 4 = m (x - 3)
All we need to do is find out what the slope of the line is. The way that we do this is by considering secant lines which intersect the circle close to the point P:
![[A series of secants PQ to a circle centred at O]](secant.gif)
Notice that as the second point of intersection gets closer and closer to P, the slope of the secant gets closer and closer to the slope of the tangent line. If we can find the slope of a general secant line, we can then use that information to approximate the slope of the tangent line.
The equation of the circle near P is
y = (25 - x2)-1/2
so if we have a point Q on the circle whose x-coordinate is 3 + h, the corresponding y-coordinate is
(25 - (3 + h)2)-1/2 = (16 - 6h - h2)-1/2
Knowing this, we can see that the slope of the secant line must be
mPQ = ((16 - 6h - h2)-1/2 - 4)/((3 + h) - 3)
Which simplifies to
mPQ = ((16 - 6h - h2)-1/2 - 4)/h
Now, the closer h is to 0, the closer Q is to P, and the closer the slope of the secant line is to the slope of the tangent line. We'd like to just substitute h = 0 into the formula, but that would evaluate to 0/0 which is undefined. Instead we plug in smaller and smaller values and look at what happens to the value.
| h | Slope |
|---|---|
| 1.0 | -1.0 |
| 0.5 | -0.858571571457 |
| 0.25 | -0.801315846429 |
| 0.1 | -0.769909508134 |
| 0.01 | -0.751956798784 |
| 0.001 | -0.750195349132 |
| 0.0001 | -0.750019531619 |
| 0.000001 | -0.750000195282 |
| 0.000000001 | -0.750000062055 |
It looks like the slope is getting closer and closer to -0.75 (or -3/4), so given this trend, we'd guess that the correct slope of the tangent line is -3/4, and in fact this is true. So the equation of the tangent line is
y - 4 = -3/4 (x - 3)
or
y = -3/4 x + 25/4
These two examples are examples of the two main sorts of problems that calculus was designed to solve - finding slopes (the differential calculus) and finding areas (the integral calculus). What links them thematically is the idea that we can approximate values that we cannot calculate directly, and use those approximations to find the precise value. The name for this procedure is taking a limit.
In the first example, we had a formula which depended on the number of sides n of the polygon and we looked at what happens as n gets very large. In that case we would express our approximation as saying that the limit of n cos(π/n) sin(π/n) as n tends to infinity is π.
In the second example, we had a formula for the slope mPQ which depended on h and we looked at what happens as h gets close to 0. In this case we would say that the limit of ((16 - 6h - h2)-1/2 - 4)/h as h tends to 0 is -3/4.
In the first part of this course we will learn how to calculate limits so we can eliminate the guesswork we used in the above examples. In the second part of this course we look at the second type of problem in detail and develop a way of calculating slopes which avoids using limits directly. In the last part of the course we investigate the area problem and discover that we can avoid limits there too, and in fact the problem turns out to be the "opposite", in a sense, to the problem of finding slopes.